World organisation for animal health (oie) Interview Questions, Process, and Tips

22

They’ll never meet unless some of them stop and other continue to run

4×1 = 4
4×2=8
8×3=24
24×4= 96 (not 92)
96×5 =480
480×6= 2880

So answer is 92

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0L -> 1 way
1L -> 3 ways
2L -> 7 ways
3L -> 4 ways
4L -> 1 way
total 16 ways

23.5 days

P Q R S

TP -> UQ -> RV -> SW

Answers-

U sits opposite to S

Q sits in between S and R (after interchange)

Ratio is 3:2:1

Total=6

Reverse of ratio is 1/3:1/2:1/1

Portion of first part is=1/3*6=>2

Portion of second part is=1/2*6=>3

Portion of third part is 1*6=>6

∴So the new ratio is 2:3:6

1/5

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

A = B + 2
B = 2 C
A + B + C = 27
C = 27 – A – B
B = 2*27 – 2A – 2B
2A = 2*27 – 3B
-2*A =-2*2 – 2*B
B = 10

2.01%

15%

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